Class 12th
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New answer posted
4 months agoContributor-Level 10
E? = φ + K?
E? = φ + K?
E? - E? = K? - K?
Now V? /V? = 2
K? /K? = 4; K? = 4K?
Now from equation (2)
⇒ 4 - 2.5 = 4K? - K?
1.5 = 3K?
K? = 0.5eV
Now putting this
Value in equation (2)
2.5 = φ + 0.5eV
φ = 2eV
New answer posted
4 months agoContributor-Level 10
Let us assume the potential at A = V_A = 0.
Now at junction C, according to KCL
i? + i? = i?
1 A + i? = 2 A
i? = 2 A
Now analyse potential along ACDB
V_A + 1 + i? (2) - 2 = V_B
0 + 1 + 2 (1) - 2 = V_B
V_B = 3 - 2
V_B = 1amp.
New answer posted
4 months agoContributor-Level 10
x = 1/√ (µ? ε? ) = speed ⇒ [x] = [L¹T? ¹]
y = E/B = speed ⇒ [y] = [L¹T? ¹]
z = l/ (RC) = l/τ ⇒ [z] = [L¹T? ¹]
So, x, y, z all have the same dimensions.
New answer posted
4 months agoContributor-Level 10
Till input voltage reaches 4 V. No zener is in breakdown region. So V? = V? then now hen V? changes between 4 V to 6 V one zener with 4 V will breakdown are P.D. across this zener will become constant and remaining potential will drop acro resistance in series with 4 V zener.
Now current in circuit increases Abruptly and source must have an internal resistance due to which. Some potential will get drop across the source also so correct graph between V? and t will be
We have to assume some resistance in series with source.
New answer posted
4 months agoContributor-Level 9
Number of bond between sulphur and oxygen
Number of bond between sulphur and sulphur
New answer posted
4 months agoContributor-Level 10
Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3
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